a, b, c are positive integers such that a<sup>2</sup> + 2b<sup>2</sup> - 2bc = 100 and 2ab - c<sup>2</sup> = 100. Then the value of <span class="math-tex">\(\dfrac{a+b}{c}\)</span> is
Subtract the second equation from the first:
\(a^2 + 2b^2 - 2bc - 2ab + c^2 = 0\)
Rearrange: \((a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) = 0\), so \((a-b)^2 + (b-c)^2 = 0\).
Both squares must be zero, so \(a = b = c\).
Then \(2a^2 - a^2 = 100\), so \(a = 10\), and \(\frac{a+b}{c} = \frac{20}{10} = 2\).
Answer: (C) 2
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